Question 187: Ray Optics

Question

In an experiment for determination of refractive index of glass of a prism by $i - \delta ,$ plot it was found thata ray incident at angle ${35^ \circ }$, suffers a deviation of ${40^ \circ }$ and that it emerges at angle ${79^ \circ }.$ In that case which of the following is closest to the maximum possible value of the refractive index?

Accepted Answer

We know that $i + e - A = \delta $
${35^ \circ } + {79^ \circ } - A = {40^ \circ }$
$	herefore$ $A = {74^ \circ }$
But $\mu  = {{\sin \left( {{{A + {\delta _m}} \over 2}} ight)} \over {\sin A/2}} = {{\sin \left( {{{74 + \delta } \over 2}} ight)} \over {\sin {{74} \over 2}}}$
$ = {5 \over 3}\sin \left( {{{37}^ \circ } + {{{\delta _m}} \over 2}} ight)$
${\mu _{\max }}\,$ can be ${5 \over 3}.$ That is ${\mu _{\max }}$ is less than ${5 \over 3} = 1.67$
But ${\delta _m}$ will be less than ${40^ \circ }$ so
$\mu  < {5 \over 3}\sin \,{57^ \circ } < {5 \over 3}\,\,$ $\sin {60^ \circ } \Rightarrow \mu  = 1.45$

Answer

$1.7$

Answer

$1.8$

Answer

$1.5$

Answer

$1.6$

Ray Optics

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